Wednesday, October 10, 2012

Cool Math Olympics - 1959 Key

1. Solution: `GCD` of `a` and `b` is the greatest common divisor of `a` and `b`. It is known that, what divides `a` and `b` also divides `a - b`. So, `GCD` of `a` and `b` should be same as that of `a` and `a - b` (when `a > b`) or that of `b` and `a - b`. Considering that, and also from the observation that, some rational number is irreducible, iff, the numerator and denominator have a `GCD` of `1`, it follows that:

Now, calculate GCD normally, like we do, divide take remainder and repeat the procedure with the remainder and the divider. The remainder in `21n + 4` when divided by `7n + 1` is `1`. which means the `GCD` is `1`. It immediately follows that, the GCD of original numbers is also `1`, hence, those are irreducible.

2. Square the expression on both sides, noting that `A >= 0` (since square root is a function, and results of that function are always non-negative).

But here, notice that the expression inside square roots is nothing but (`x^2 - 2x + 1`, which is a square expression). Now, again;

`2x + 2|x - 1| = A^{2}, (A >= 0, x >= 1/2)` <=>

(`|x|` is the absolute value of `x`).
The condition `x >= 1/2` is because of `\sqrt{2x - 1}`, where the one inside square-root cant be negative.

Now substitute values for `A`,

(i) `A = 1`, `x + |x - 1| = 1/2`.

Notice that, `|x - 1|` cant be `1 - x`. If it is so, `A^{2}/2 = 1` always, regardless of the actual value of A. So, this yields another condition:

If `A^{2} / 2 != 1`, then `x >= 1`.

Consider these conditions into the expression (i), and you get `x = 3/4`, which is not according to the conditions. So, (i) does not have a solution.

(ii) `A = 1/2`, `x + |x - 1| = 1/8`.

Here, we get `x = 9/16`, since this is also less than `1`, (ii) also has no solution.

(iii) `A = 2`, `x + |x - 1| = 2`. Here you get, `x = 3/2`, which satisfies all conditions. So, only this combination has solutions for `x`, which is given by `3/2`.

3. Remember `cos(2x) = 2cos^{2}(x) - 1`. And, in a quadratic equation `ax^{2} + bx + c = 0`, the sum of roots is given by `-b/a` and product is given by `c/a`. First, derive formulae for `cos(2x_{1}) + cos(2x_{2}) ` and `cos(2x_{1})cos(2x_{2})` in terms of `cos(x_{1}) + cos(x_{2})` and `cos(x_{1})cos(x_{2})`, where `x_{1}` and `x_{2}` are roots of original equation.

When you substitute, `a = 4, b = 2, c = -1`, you get. the roots for original equation:

`\alpha = (-1 + \sqrt{5}) / 4, and \beta = (-1 - \sqrt{5}) / 4`.

Surprisingly, the equation in `cos(2x)` also yields the same roots. and in fact, the sane equation when simplified by eliminating common factors between co-efficients.

However, there is no need to be surprised. Because, calculate `cos(2x)` for `\alpha`, it will be `\beta` and `cos(2x)` for `\beta` will be `\alpha`. So, both the equations are right, and compatible. (Its a neat trick substitution).

4. Consider an angle in the right triangle to be `\theta`, where `\theta != 90^{o}`. The median divides the hypotenuse into 2 equal lengthed parts. Since the lengh of hypotenuse is `c`, one side will be `c * cos(\theta)` and another will be `c * sin(\theta)`. The median divides the right triangle into two other triangles. Consider the triangle inside which we have angle `\theta`. Using the cosine rule of lengths of sides, gives the following:

(To brush up: Cosine rule says that, if angle between sides length `a` and `b` is `\theta`, the third side in the triangle is given by `sqrt{a^{2} + b^{2} - 2ab * cos(\theta)}`).

Basically, to construct such a triangle, with any `c`, choose an angle to be `15^{o}` in the right triangle.

5. Now to the last problem, this is a geom problem. I will present anal. geom. solution except for the last part. (Which is too easy in co-ordinate geometry).

Consider this theorem, well known in circles of (pun is intended), ananlytical geom. If a Jyaa, `AB` in a circle makes an angle `\theta` at the center `C` of the circle, (namely, `<ACB` is `\theta`), then on any point on the circle, to the side of `C`, it makes an angle of `< \theta / 2`). With the help of this theorem, you can easily solve this problem.

First of all, notice that `<ANM` and `<MNB` are both `45^{o}`, since `<APM` and `<MQB` are `90^{o}` (`P` and `Q` are also centers of circum-circle for a square).

(ii) Consider for now, `A` is the origin, and `AB` is the `x` axis. Consider `AM = l_{1}` and `AB = l`. Consider the point `R` at `(l/2, -l/2)`. The angle `<BMR = 135^{o} - \theta` since the opposite angle `<AMN` is of same measure.